剑指Offer之Java算法习题精讲二叉树的构造和遍历

题目一

 解法

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public TreeNode constructMaximumBinaryTree(int[] nums) {        return method(nums,0,nums.length-1);    }    public TreeNode method(int[] nums,int lo,int hi){        if(lo>hi){            return null;        }        int index = -1;        int max = Integer.MIN_VALUE;        for(int i = lo;i<=hi;i++){            if(max<nums[i]){                max = nums[i];                index = i;            }        }        TreeNode root = new TreeNode(max);        root.left = method(nums,lo,index-1);        root.right = method(nums,index+1,hi);        return root;    }}

题目二

 解法

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        return method(preorder,0,preorder.length-1,inorder,0,inorder.length-1);    }    public TreeNode method(int[] preorder, int preLeft,int preEnd , int[] inorder,int inLeft,int inEnd){        if(preLeft>preEnd){            return null;        }        int rootVal = preorder[preLeft];        int index = -1;        for(int i = inLeft;i<=inEnd;i++){            if(rootVal == inorder[i]){                index = i;            }        }        TreeNode root = new TreeNode(rootVal);        int leftSize = index - inLeft;        root.left = method(preorder,preLeft+1,leftSize+preLeft,inorder,inLeft,index-1);        root.right = method(preorder,leftSize+preLeft+1,preEnd,inorder,index+1,inEnd);        return root;    }}

题目三

 解法

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public TreeNode buildTree(int[] inorder, int[] postorder) {        return build(inorder,0,inorder.length-1,postorder,0,postorder.length-1);    }    TreeNode build(int[] inorder, int inStart, int inEnd,int[] postorder, int postStart, int postEnd) {    if (inStart > inEnd) {        return null;    }    // root 节点对应的值就是后序遍历数组的最后一个元素    int rootVal = postorder[postEnd];    // rootVal 在中序遍历数组中的索引    int index = 0;    for (int i = inStart; i <= inEnd; i++) {        if (inorder[i] == rootVal) {            index = i;            break;        }    }    // 左子树的节点个数    int leftSize = index - inStart;    TreeNode root = new TreeNode(rootVal);    // 递归构造左右子树    root.left = build(inorder, inStart, index - 1,postorder, postStart, postStart + leftSize - 1);    root.right = build(inorder, index + 1, inEnd,postorder, postStart + leftSize, postEnd - 1);    return root;}}

题目四

 解法

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode() {} *     TreeNode(int val) { this.val = val; } *     TreeNode(int val, TreeNode left, TreeNode right) { *         this.val = val; *         this.left = left; *         this.right = right; *     } * } */class Solution {    public TreeNode constructFromPrePost(int[] preorder, int[] postorder) {        return method(preorder,0,preorder.length-1,postorder,0,postorder.length-1);    }    public TreeNode method(int[] preorder,int preStart, int preEnd, int[] postorder,int postStart,int postEnd){        if(preStart>preEnd){            return null;        }        if(preStart==preEnd){            return new TreeNode(preorder[preStart]);        }        int rootVal = preorder[preStart];        int leftRootVal = preorder[preStart + 1];        int index = 0;        for (int i = postStart; i < postEnd; i++) {            if (postorder[i] == leftRootVal) {                index = i;                break;            }        }        TreeNode root = new TreeNode(rootVal);        int leftSize = index - postStart + 1;        root.left = method(preorder, preStart + 1, preStart + leftSize,postorder, postStart, index);        root.right = method(preorder, preStart + leftSize + 1, preEnd,postorder, index + 1, postEnd - 1);        return root;    }}

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